I am trying to prove the following statement:

"If $B\subset\mathbb{R}$ is a Borel set and $f:B\to\mathbb{R}$ is an increasing function then $f(B)$ is a Borel set"

but I have only managed to prove the easier statement

"If $B\subset\mathbb{R}$ is a Borel set and $f:\mathbb{R}\to\mathbb{R}$ is a strictly increasing function then $f(B)$ is a Borel set"

My proof (of the easier statement):

By Inverse function $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ of a strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ is continuous we have that $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ is continuous so $(f^{-1})^{-1}(\mathbb{R})=f(\mathbb{R})$ is open hence Borel thus $f^{-1}$ is a continuous function defined on a Borel set so it is Borel measurable which implies that $(f^{-1})^{-1}(B)=f(B)$ is a Borel set, as desired. $\square$

I would like to prove the initial statement but I have been stuck for a while so I would appreciate an hint about how to tackle its proof, thanks.


1 Answer 1


Here is a self-contained proof. We first note that it suffices to prove:

Claim. Let $f : \mathbb{R} \to \mathbb{R}$ be non-decreasing. Then for each Borel set $B \subseteq \mathbb{R}$, the set $f(B)$ is also Borel.

To prove this, let $\mathcal{F}$ be the set of all Borel sets $B$ for which $f(B)$ is also Borel. Then we check the following properties of $\mathcal{F}$:

1. $\mathcal{F}$ contains any open intervals.

Indeed, let $I$ be an open interval, and consider any connected component $C$ of $\mathbb{R}\setminus f(I)$.1) We show that $C$ cannot be a singleton. Otherwise, write $C = \{y\}$. Then we can find sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ in $I$ such that

\begin{gather*} f(a_1) < f(a_2) < \cdots < y < \cdots < f(b_2) < f(b_1), \\ \lim f(a_n) = y, \qquad \lim f(b_n) = y. \end{gather*}

This in particular forces that $a_1 < a_2 < \cdots < b_2 < b_1$, and so, both $a = \lim a_n$ and $b = \lim b_n$ exist in $I$ and $a \leq b$. Moreover, $y \leq f(a) \leq f(b) \leq y$, and so, $f(a) = f(b) = y$. This contradicts that $y$ lies in the complement of $f(I)$, and so, $C$ is not singleton as required.

Now this tells that $\mathbb{R}\setminus f(I)$ is a disjoint union of nondegenerate intervals, and since there are at most countably many such intervals, their union is a Borel set. Therefore $f(I)$ is also a Borel set.

2. If $A_1, A_2, \dots$ are in $\mathcal{F}$, then both $\cup_n A_n$ and $f(\cup_n A_n) = \cup_n f(A_n)$ are Borel sets, hence $\cup_n A_n \in \mathcal{F}$.

3. If $A \in \mathcal{F}$, then $\mathbb{R}\setminus A \in \mathcal{F}$.

Indeed, denote by $D$ the set of values in $\mathbb{R}$ which are attained by $f$ at two or more points, i.e.,

$$ D = \{y \in \mathbb{R} : \#f^{-1}(\{y\}) \geq 2. \} $$

For each $y \in D$ and for each $a, b \in f^{-1}(\{y\})$ with $a < b$, we find that $[a, b] \subseteq f^{-1}(\{y\})$. This tells that each $f^{-1}(\{y\})$ is a nondegenerate interval, and so, $D$ can contain at most countably many points. Then by noting that $f(A) \cap f(\mathbb{R}\setminus A) \subseteq D$, we have

$$ f(\mathbb{R})\setminus f(A) \subseteq f(\mathbb{R}\setminus A) \subseteq (f(\mathbb{R})\setminus f(A)) \cup D. $$

In particular, $f(\mathbb{R}\setminus A) = (f(\mathbb{R})\setminus f(A)) \cup \tilde{D}$ for some subset $\tilde{D}$ of $D$. Since all of $f(\mathbb{R})$, $f(A)$, and $\tilde{D}$ are Borel sets, $f(\mathbb{R}\setminus A)$ is also a Borel set and hence $\mathbb{R}\setminus A \in \mathcal{F}$ as desired.

Conclusion. The above observations tell that $\mathcal{F}$ is a $\sigma$-algebra containing all the open intervals, and hence, must contain any Borel sets. Therefore the proof is complete. $\square$

1) If you are not familiar with the notion of connect components, then here is an alternative way of explaining this for subsets of $\mathbb{R}$. Let $E \subseteq \mathbb{R}$, and define the relation $\sim$ on $E$ as follows:

  • For each $x, y \in E$, we write $x \sim y$ if and only if $\{tx + (1-t)y : t \in [0, 1]\} \subseteq E$.

In other words, $x \sim y$ if and only if either $x = y$ or the closed interval between $x$ and $y$ lie in $E$. Then it is not hard to check that

  1. $\sim$ is an equivalence relation on $\mathbb{R}$, and
  2. each equivalence class of $\sim$ is either a singleton or an interval.
  • $\begingroup$ thank you very much, I have a few questions about your proof; since I am not very familiar with topology could you please tell me why in point (1) you consider a "connected component"? What is it and what is the intuition behind this choice? Also, shouldn't the inequality in (1) be $f(a)\leq y\leq f(b)$? Also, in part (3) you say "For each $y\in D$ and for each $a,b\in f^{-1}(\{y\})$ with $a<b$, we find that $[a,b]\subseteq f^{-1}(\{y\})$": why is that? Thanks $\endgroup$
    – lorenzo
    Commented Aug 2, 2021 at 14:07
  • 1
    $\begingroup$ @lorenzo, For (1), I added a brief explanation to my answer. The idea is that if $f$ is increasing and makes a jump at a point, then that jump will manifest as intervals in the complement $\mathbb{R}\setminus f(I)$. For instance, if we consider $f(x)=\operatorname{sgn}(x)$, then $$\mathbb{R}\setminus f((-1,1))=(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty),$$ and the part $(-1,0)\cup(0,1)$ corresponds to the jump of $\operatorname{sgn}(x)$ at $x=0$. $\endgroup$ Commented Aug 2, 2021 at 17:54
  • 1
    $\begingroup$ @lorenzo, for (2), the inequality you mentioned is also true, but it alone does not nail down the values of $f(a)$ and $f(b)$. To explain why we expect $y\leq f(a)\leq f(b)\leq y$ to hold, the idea is that $C$ cannot be a singleton because $C$ corresponds to (at least part of) a jump of $f$, and the zero jump size will imply continuity, i.e., no jump. Now using $(a_n)$ and $(b_n)$ chosen as in the answer, note that $a_n \leq a \leq b \leq b_n$, and so, $f(a_n) \leq f(a) \leq f(b) \leq f(b_n)$. Then letting $n\to\infty$ and $\lim f(a_n)=\lim f(b_n) = y$ will give us the inequality. $\endgroup$ Commented Aug 2, 2021 at 17:59
  • 1
    $\begingroup$ @lorenzo, Finally, the claim mentioned in (3) more or less tells that if you have $f(a) = f(b)$, then the monotonicity of $f$ forces that the graph of $f$ on the interval $[a, b]$ must be flat, i.e., $f(x) = f(a) = f(b)$ for all $x \in [a,b]$. This then forces that $[a,b]$ lies in the inverse image of $\{y\}$ under $f$. $\endgroup$ Commented Aug 2, 2021 at 18:00
  • $\begingroup$ fantastic answer! Thank you very much! $\endgroup$
    – lorenzo
    Commented Aug 3, 2021 at 14:37

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